show-that-i-i-is-always-real-

Question Number 151560 by peter frank last updated on 21/Aug/21

show that i^{i} is always real

Answered by puissant last updated on 21/Aug/21

i^{2} = - 1 \Rightarrow i = \sqrt{- 1}

i^{i} = {(\sqrt{- 1})}^{i} = e^{i l n (\sqrt{- 1})} = e^{\frac{i}{2} {l n i}^{2}} = e^{i l n i}

= e^{i \frac{π}{2} i} = e^{- \frac{π}{2}} \in R . .

Commented by peter frank last updated on 22/Aug/21

thank you

Answered by Olaf_Thorendsen last updated on 21/Aug/21

z = i^{i}

\ln z = i \ln i = i \ln (e^{i \frac{π}{2}}) = i (i \frac{π}{2}) = - \frac{π}{2}

z = e^{- \frac{π}{2}} \in R .

Commented by peter frank last updated on 22/Aug/21

thank you

Answered by MJS_new last updated on 21/Aug/21

i^{i} = {(e^{i \frac{π}{2}})}^{i} = e^{i^{2} \frac{π}{2}} = e^{- \frac{π}{2}}

Commented by peter frank last updated on 22/Aug/21

thank you

Answered by peter frank last updated on 22/Aug/21